Warmup Problem

{ a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b)\\ { 99 }^{ 2 }-{ 98 }^{ 2 }=(99+98)(99-98)\\ \quad \quad \quad \quad =(197)(1)\\ \quad \quad \quad \quad =\boxed { 197 }

$99^2 -98^2= (98+1)(98+1)-98^2$ $=98^2+98+98+1-98^2=2*98+1=197$

n^2 - (n-1)^2 = 2n-1

Imagine a 98x98 square fitting inside of a 99x99 square. The remaining open space is the difference, which yields a row of 99, a column of 99, and an overlap. Thus 99+99-1=197

It's been a while since college algebra, so: mentally, 100^2=10,000. 99^2 is one 100 and one 99 less than that: 9,801. 98^2 is two 100's and two 98's less than 10,000: 9,604. The difference would be 200 if it weren't for the fact that 4 is 3 more than 1; therefore, the difference is 3 less than 200.

99^2 - 98^2 = (98+1)^2 - 98^2 = (a+1)^2 - a^2 = a^2 - a^2 + 2a + 1= 2*98 + 1 = 196 +1 = 197

$99^2-98^2=(99+98)(99-98)=(197)(1)=\boxed{197}$

A pattern of squares is that

a^2=b^2+(a+b)

Where a is the next number and b is the previous

99²-98²=

(98+1)(98+1)-98²=

98²+2(98)+1²-98²=

2(98)+1= 197

Difference of two squares formula: 99^2 - 98^2= (99 + 98)(99 - 98)= (197)(1)= 197

I did it by noticing the pattern of the difference in powers. I noticed that if I did

$3^{2} - 2^{2} = 5$

$4^{2} - 3^{2} = 7$

$5^{2} - 4^{2} = 9$

And so on. So I realized that the given the format $a^{2} - b^{2}$ the answer was always $(b \times 2) + 1$

From there it was as easy as $(98 \times 2) + 1 = 197$

99~2-98~2 =(99+98) (99-98)=197X1= 197Write a solution.

$99^2-98^2 = (98+1)^2 - 98^2$

$\quad \quad \quad \quad = 98^2 + 2\times 98 + 1^2 - 98^2$

$\quad \quad \quad \quad = 196 + 1 = \boxed {197}$

99^2-98^2=9801-9604=197

99^2 - 98^2=(99+98) (99-98) = 197 (1) = 197

I forgot about difference of two squares, so I did this:

99^2 = (98+1)(98+1) = 98^2 + 98 + 98 + 1

so

99^2 - 98^2 = 98^2 + 98 + 98 + 1 - 98^2 = 98 + 98 + 1 = 197

Best way to work this out is visually, Imagine a 99 X 99 pixel grid then a 98 X 98 pixel grid on top of it, there will two 1-pixel lines forming an L-shape which give the answer, The lines are each 99*1 but with one common pixel. ie. ((99 X 1) X 2)-1

As a matter of pattern, if (2^2)-(1^2) = 3 and 1+2=the same, and (5^2)-(4^2)= 9and 5+4 =the same, and (7^2)-(6^2)=49-36=13=7+6, and so one, a pattern can be seen that a number squared minus one less than itself squared is equivalent to both numbers added, therefore, simply though pattern recognition, without actually having to actually "do" the equation itself, we can recognize that (99^2)-(98^2) is equivalent to 99+98, in other words 197. So yes, a calculator would have taken the fun out of that.

Difference between the squares of two consecutive numbers will be the sum of the two numbers always.

(a+1)^2 -a^2 = 2a+1

99^2 - [(99-1)^2]

99^2 - 98^2 =99 99 -98^2 =(98+1)(98+1) -98^2 =98^2+2 98 +1 -98^2 =2*98 +1 =197

99^2 = (99 * 99)-(98^2) 99 * 99 = (98 + 1) (98 +1) - 98 98 = [(98 * 98) + (98 * 1) + (98 * 1) + (1 * 1)] - (98 98) =98+98+1 =197

(a+b)^2 = (a+b)(a-b), so it becomes 98+ 99 = 197

X^2-(X-1)^2 = X+(X-1) 99^2-98^2 =99+98=197

We use concept: a^2-b^2 = (a+b) (a-b) So, Obviously that 99^2 - 98^2 = (99+98)(99-98) = 197.

I just did 99 + 98

Difference of two squares results in 197.

(Sequential exponential pattern)

n=1 n^2=1 (n+1)^2 = n^1 +n x 2 +1 4 = 1 + 2 + 1 4=4

sub in 98 n^2-n^2+ n x 2 + 1 = the difference of the two squares which is 197

First, you should know the pattern of difference of two square numbers:

a^2 - b^2 = (a+b) (a-b)

Just follow the pattern/formula and you will get the answer.

x^2-y^2= x+y iff x=y+1

99^2-98^2=(98+1)(98+1)^2-98^2 =98^2+98+98+1-98^2=98+98+1=196+1=197

Why dont we just use easiest solution? 99^2 - 98^2 = 99 + 98 = 197

square of a number s d sum of preceding square nd its double +1 ....... i.e.......2^2 = 4 .......3^2= 4+(2 2)+1..... (3^2)-(2^2) = (2 2)+1... ......in d same way ....(99^2) - (98^2) = (98*2)+1...

If n^2 is known, (n+1)^2 = n^2 + 2n + 1

The nth term for the differences between squares is 2n+1 . The question asks for the difference between the 98th and 99th squares, so we substitute 98 as n to find the 98th difference.

(2)98+1 = 197

Let a=98. Then 99^2 - 98^2 = (a+1)^2 - a^2 = a^2 + 2a + 1 - a^2 = 2a + 1 = 2a+1 = 2(98) + 1 = 197

A Solution for the absolute diferance of any 2 consecutive squares

(99+98)(99-98)=197

(100-1)^2-(100-2)2= 100^2+1-200-100^2-4+400
=197

let x = 99, then x^2 - (x-1)^2 = x^2 - (x^2-2x+1)=2x-1=2*99 - 1 = 197 =D

(a^2)-(b^2)=(a+b)(a-b) =(99^2)-(98^2)=(99+98)(99-98)=197

Write a solution.

a^2 - b^2 = a + b

I just added 99 and 98 together. Is it logical to u guys?

99^2 = 99 * 99 = 9801 ;98^2 = 98 * 98 = 9604 ;9801-9604 = 197

99^2 - 98^2 = (98+1)^2-98^2 = (98^2 + 2*98 +1) - 98^2 = 197

(100-1)^2-(100-2)^2=(100^2-200+1-100^2+400-4)=197

99^2 - 98^2

= (100-1)^2 - (100-2)^2 - Step 1

Using the formula (a-b)^2 = a^2-2ab+b^2, we can compute step 2 to be

9801 - 9604 = 197

(98+1)^2-(98^2)= (98^2)+98+98+1-(98^2)

99²-98²=197 (99x99)-(98x98) =197 9801-9604 = 197

SIMPLE SOLUTION ^^

Simple if it is square of both then the difference is the sum of them.Like 99²-98²=99+98 =197

99^2-98^2 =(98+1)^2-98^2 =98^2+1^2+2 98 1-98^2 =1+196 =197

99^2=98^2+98+99 so 99^2-98^2=98+99=197

If a=2 and b=1, then

Then

So

Ritu Roy knows whatsup

The difference between squares of two positive consecutive numbers will always be the sum of those two numbers. So 99^2-98^2=99+98=197

99^{2} - 98^{2} = (98 + 1)^{2} - 98^{2} = (98^{2} +2 98+1) - 98^{2} = 2 98 + 1 = 197

in this way too, we can solve it (100-1)^2 - (100-2)^2 by using (a-b)^2 formula we can solve

a^2-b^2 = (a+b)(a-b) 99^2 - 98^2 = (99+98)(99-98) =197(1) = 197

a^2-b^2=(a+b)(a-b) 99^2-98^2=(99+98)(99-98) =197*1 =197

Logic/geometric solution : Draw a square with a 98 side into a 99 side one : result of exclusion is a horizontal stripe 99×1 and a vertical stripe 98×1. So answer is 98+99=197

99^2 - 98^2 = 99^2 - (99 - 1)^2 = 99^2 - (99^2 - 2 99 + 1) = 2 99 -1 = 197

98+99=197 bcoz difference b/w 2 consecutive squares is the sum of those 2 consecutive numbers

99 ka square ly gy nd phr usy 98 k square sy (-) minous kr dy gy :)

(99+98)(99-98) (197)(1) 197

98^2=(99-1)^2

99^2-(99^2-2x99+1)

=2x99-1=197

a^-b^=(a+b)(a-b)
99^-98^ =(99+98)(99-98)

(100-1)^{ 2 }-(100-2)^{ 2 }=((100^{ 2 })-(2 100 1)+1)-((100^{ 2 })-(2 100 2)+(2^{ 2 }))=400-200-4+1=197

(1+98)(1+98)-98^2 =1+2×98+98^2-98^2 =1+196 =197

99^2-98^2 = (100-1)^2 - (100-2)^2 = (10,000-200+1) - (10,000-400+4) =10,000-200+1-10,000+400-4 =200-3 =197

Difference of two squares formula: 99^2 - 98^2= (99 + 98)(99 - 98)= (197)(1)= 197

99^2 - 98^= 9801 - 9604=197

99^2-98^2=(98+1)^2-98^2=98^2+2 98 1+1-98^2=196+1=197

99^2 -98^2 =(99+98) =197